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🚩第1次作业
1.写出下列随机试验的样本空间及下列事件包含的样本点.
涉及到的概念：
随机试验E的所有基本结果组成的集合称为样本空间（sample space），记为Ω.样本,"> 
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            <h1 id="概率论与数理统计作业"><a href="#概率论与数理统计作业" class="headerlink" title="概率论与数理统计作业"></a>概率论与数理统计作业</h1><hr>
<h2 id="🚩第1次作业"><a href="#🚩第1次作业" class="headerlink" title="🚩第1次作业"></a>🚩第1次作业</h2><hr>
<p>1.写出下列随机试验的样本空间及下列事件包含的样本点.</p>
<pre><code class="txt">涉及到的概念：
随机试验E的所有基本结果组成的集合称为样本空间（sample space），记为Ω.样本空间的元素，即E的每个基本结果，称为样本点.</code></pre>
<p>（1） 掷一颗骰子，出现奇数点.</p>
<p>解：<br>$$<br>\begin{align}<br>&amp;\Omega_1={1,2,3,4,5,6 };\<br>&amp;样本点为{1,3,5}<br>\end{align}<br>$$<br>（2） 掷二颗骰子，</p>
<p>  <em>A</em>=“出现点数之和为奇数，且恰好其中有一个1点.”</p>
<p>  <em>B</em>=“出现点数之和为偶数，但没有一颗骰子出现1点.”</p>
<p>解：<br>$$<br>\Omega_2={(i,j)|i,j=1,2,3,4,5,6};<br>$$</p>
<p>$$<br>A={(1,2),(1,4),(1,6),(2,1),(4,1),(6,1)}<br>$$</p>
<p>$$<br>B={(2,2),(4,2),(6,2),(2,4),(4,4),(6,4),(2,6),(4,6),(6,6),}<br>$$</p>
<p> （3）将一枚硬币抛两次，</p>
<p> <em>A</em>=“第一次出现正面.”</p>
<p><em>B</em>=“至少有一次出现正面.”</p>
<p><em>C</em>=“两次出现同一面.” </p>
<p>解：<br>$$<br>\Omega_3={(正,反),(正,正),(反,正),(反,反)}<br>$$</p>
<p>$$<br>A={(正,反),(正,正)}<br>$$</p>
<p>$$<br>B={(正,反),(正,正),(反,正)}<br>$$</p>
<p>$$<br>C={(正,正),(反,反)}<br>$$<br>2.设<em>A</em>，<em>B</em>，<em>C</em>为三个事件，试用<em>A</em>，<em>B</em>，<em>C</em>的运算关系式表示下列事件：</p>
<p>（1） <em>A</em>发生，<em>B</em>，<em>C</em>都</p>
<p>（2） <em>A</em>，<em>B</em>，<em>C</em>都发生； </p>
<p>（3） <em>A</em>，<em>B</em>，C至少有1个发生；</p>
<p>（4） <em>A</em>，<em>B</em>，<em>C</em>都不发生； </p>
<p>（5） <em>A</em>，<em>B</em>，<em>C</em>不都发生；</p>
<p>（6） <em>A</em>，<em>B</em>，<em>C</em>至多有1个不发生. </p>
<p>解：<br>$$<br>(1)A\overline B\overline C<br>$$<br>$$<br>(2)ABC<br>$$<br>$$<br>(3)A\cup B\cup C=\overline{AB}C\cup \overline AB\overline C\cup A\overline B\overline C\cup \overline ABC\cup A\overline BC\cup AB\overline C\cup ABC=\overline{\overline{AB}\overline C}<br>$$<br>$$<br>(4)\overline A\overline B\overline C=\overline{A\cup B\cup C}<br>$$<br>$$<br>(5)\overline{ABC}<br>$$<br>$$<br>(6)ABC\cup \overline ABC\cup A\overline BC\cup AB\overline C<br>$$</p>
<pre><code class="txt">e.g.:
例1.1  设A，B，C为3个事件，试用A，B，C的运算式表示下列事件：
（1） A发生而B与C都不发生；
（2） A，B都发生而C不发生；
（3） A，B，C至少有两个事件发生；
（4）A，B，C至多有两个事件发生；
（5） A，B，C恰有两个事件发生：
（6） A，B中至少有一个发生而C不发生.
解
（1）A¯B ¯C；
（2）AB¯C；
（3）（AB）∪（AC）∪（BC）；
（4）¯A∪¯B∪¯C；
（5）（AB¯C）∪（A¯BC）∪（¯ABC）； 
（6）（A∪B）¯C.</code></pre>
<hr>
<h2 id="🚩第2次作业"><a href="#🚩第2次作业" class="headerlink" title="🚩第2次作业"></a>🚩第2次作业</h2><hr>
<p><img src="https://gitee.com/L3SLIE/blog/raw/master/img/%E6%A6%82%E7%8E%87%E8%AE%BA%E4%B8%8E%E6%95%B0%E7%90%86%E7%BB%9F%E8%AE%A1%E7%AC%AC%E4%B8%80%E7%AB%A0%E7%AC%AC%E4%B8%89%E9%A2%98.png" alt="习题3"></p>
<p>解：</p>
<ol>
<li><p>不成立；<br>$$<br>\begin{align}<br>若 A\cap B=\varnothing,则AB\cup B=B<br>\end{align}<br>$$</p>
</li>
<li><p>不成立；<br>$$<br>若事件A发生，则\overline A不发生，A\cup B发生，则\overline AB不发生<br>$$</p>
</li>
<li><p>不成立；<br>$$<br>题目？？？(\overline{A\cup B}\cap C=\overline A\overline BC)\<br>若A-B发生，则\overline{AB}发生，\overline{A\cup B}不发生，故不成立。<br>$$</p>
</li>
<li><p>成立；<br>$$<br>AB与\overline{AB}为互斥事件<br>$$</p>
</li>
<li><p>成立；<br>$$<br>若事件A发生，则事件B发生，所以AB发生；\<br>若事件AB发生，则事件A发生，事件B发生。\<br>故成立。<br>$$</p>
</li>
<li><p>成立；<br>$$<br>若事件C发生，则事件A发生，则事件B不发生，故BC=\varnothing<br>$$</p>
</li>
<li><p>不成立；<br>$$<br>\overline B \subset \overline A<br>$$</p>
</li>
<li><p>成立；<br>$$<br>A\subset(A\cup B)，若事件A发生，则事件A\cup B发生；\<br>若事件A\cup B发生，则事件A发生，事件B发生；\<br>故成立。<br>$$</p>
</li>
</ol>
<p><img src="https://gitee.com/L3SLIE/blog/raw/master/img/%E6%A6%82%E7%8E%87%E8%AE%BA%E4%B8%8E%E6%95%B0%E7%90%86%E7%BB%9F%E8%AE%A1%E7%AC%AC%E4%B8%80%E7%AB%A04%EF%BC%8C5.png" alt="习题4&amp;5"></p>
<p><strong>4.</strong><br>$$<br>解：P(\overline{AB})=1-P(AB)=1-[P(A)-P(A-B)]=1-(0.7-0.3)=0.6<br>$$<br><strong>5.</strong><br>$$<br>\begin{align}<br>解：&amp;\<br>&amp;(1)在AB=A时，P(AB)取到最大值0.6；\<br>&amp;(2)在A\cup B=\varnothing 时，P(AB)取到最小值0。<br>\end{align}<br>$$</p>
<hr>
<h2 id="🚩第3次作业"><a href="#🚩第3次作业" class="headerlink" title="🚩第3次作业"></a>🚩第3次作业</h2><hr>
<p>p29习题一</p>
<p>//8，10，11，13，14</p>
<p><strong>8.</strong> 对一个5人学习小组考虑生日问题，求：</p>
<p>（1） 5个人的生日都在星期日的概率；</p>
<p>（2） 5个人的生日都不在星期日的概率；</p>
<p>（3） 5个人的生日不都在星期日的概率.<br>$$<br>\begin{align}<br>&amp;解：\</p>
<p>&amp;（1）设A_1={五个人生日都不在星期日}，基本事件总数为7^5，有利事件只有一个，\&amp;故P(A_1)={1\over 7^5}=({1\over 7})^5\<br>&amp;（2）设A_2={五个人生日都不在星期日}，有利事件数位6^5,\&amp;故P(A_2)={6^5\over 7^5}=({6\over 7})^5\<br>&amp;（3）设A_3={五个人生日都不在星期日}\&amp;P(A_3)=1-p(A_1)=1-({1\over 7})^5<br>\end{align}<br>$$</p>
<p>10.一批产品共<em>N</em>件，其中<em>M</em>件正品.从中随机地取出<em>n</em>件（<em>n</em>&lt;<em>N</em>）.试求其中恰有<em>m</em> 件</p>
<p>（<em>m</em>≤<em>M</em>）正品（记为<em>A</em>）的概率.如果：</p>
<p>（1） <em>n</em>件是同时取出的；</p>
<p>（2） <em>n</em>件是无放回逐件取出的；</p>
<p>（3） <em>n</em>件是有放回逐件取出的.<br>$$<br>\begin{align}<br>解：&amp;\<br>&amp;(1)P(A)=C^m_MC^{n-m}<em>{N-M}/C^n_N\<br>&amp;(2)无放回逐件取出，样本总数有P^n_N种，n次抽取中有m次为正品的组合数为C^m_M种，\&amp;对于固定的一种正品与次品的抽取次序，从M件正品中取出m件的排列数有P^m_M种，\&amp;从N-M件次品中取出n-m件的排列数为P^{n-m}</em>{N-M}<br>\&amp;P(A)=C^m_nP^m_MP^{n-m}_{N-M}/P^n_N\<br>&amp;(3)有放回，可能的取法总数为N^m种，n此抽取有m次为正品的组合数为C^m_n种，\&amp;m次取到正品都有M种取法，共有M^m种取法，n-m次取到次品，每次都有N-M种取法，\&amp;故P(A)=C^m_nM^m(N-M)^n-m/N^n</p>
<p>\end{align}<br>$$<br>11.在电话号码簿中任取一电话号码，求后面4个数全不相同的概率（设后面4个数中的每一个数都是等可能地取自0，1，…，9）.<br>$$<br>\begin{align}<br>解：&amp;后四位的可能情况有，4^{10}种，全不相同的情况有C_4^2<em>10+C_4^3</em>10+C_4^4*10，\&amp;所以全不相同的概率P(A)=1-10/4^{10}-40/4^{10}-60/4^{10}=1-110/4^{10}<br>\end{align}<br>$$<br>13.一个袋内装有大小相同的7个球，其中4个是白球，3个是黑球，从中一次抽取3个，计算至少有两个是白球的概率.<br>$$<br>\begin{align}<br>解：&amp;设A_i={恰有i个白球}(i=2,3),显然A_2与A_3互斥\<br>&amp;P(A_2)=C_4^2C_3^1/C_7^3=18/35\<br>&amp;P(A_3)=C_4^3/C_7^3=4/35\<br>&amp;P(A_2\cup A_3)=P(A_2)+P(A_3)=22/35<br>\end{align}<br>$$<br>14.有甲、乙两批种子，发芽率分别为0.8和0.7，在两批种子中各随机取一粒，求：</p>
<p>（1）两粒都发芽的概率；</p>
<p>（2）至少有一粒发芽的概率；</p>
<p>（3）恰有一粒发芽的概率.<br>$$<br>\begin{align}<br>解：&amp;设A_i={第i批种子中的一粒发芽}，(i=1,2)\<br>&amp;(1)P(A_1A_2)=P(A_1)P(A_2)=0.7\times 0.8=0.56\<br>&amp;(2)P(A_1\cup A_2)=0.7+0.8-0.7\times 0.8=0.94\<br>&amp;(3)P(A_1\overline A_2\cup \overline A_1A_2)=0.8\times 0.3+0.2\times 0.7=0.38<br>\end{align}<br>$$</p>
<hr>
<h2 id="🚩第4次作业"><a href="#🚩第4次作业" class="headerlink" title="🚩第4次作业"></a>🚩第4次作业</h2><hr>
<p>18.某地某天下雪的概率为0.3，下雨的概率为0.5，既下雪又下雨的概率为0.1,求：</p>
<p>（1）在下雨条件下下雪的概率；</p>
<p>（2）这天下雨或下雪的概率.<br>$$<br>\begin{align}<br>解：&amp;(1)雨天下雪的概率为 0.5\times0.3=0.15\<br>&amp;(2)下雨或下雪的概率为 0.5+0.3=0.8<br>\end{align}<br>$$<br>21.两人约定上午9∶00~10∶00在公园会面，求一人要等另一人0.5h以上的概率.<br>$$<br>\begin{align}<br>解：&amp;设两人到达的时刻为x，y则0\le x,y\le 60;\&amp;一人等另一人0.5h以上|x-y|&gt;30\&amp;联立解不等式得概率为{1\over 4}<br>\end{align}<br>$$</p>
<ol start="23">
<li></li>
</ol>
<p>$$<br>设P(\overline A)=0.3,P(B)=0.4,P(A\overline B)=0.5,求P(B|A\cup\overline B)<br>$$</p>
<p>$$<br>\begin{align}<br>解：&amp;P(\overline A)=0.3,P(B)=0.4,P(A\overline B)=0.5;\<br>&amp;P(A\cup B)=1-0.5=0.5;\<br>&amp;P(A)=1-0.3=0.7;P(\overline B)=1-0.4=0.6\<br>&amp;P(A\cup \overline B)=P(A)+P(\overline B)-P(A\overline B)=0.8\<br>&amp;P(AB)=P(A)-P(A\overline B)=0.2\<br>&amp;P(B|A\cup\overline B)={P(BA)\over P(A\cup \overline B)}=1/4<br>\end{align}<br>$$</p>
<hr>
<h2 id="🚩第5次作业"><a href="#🚩第5次作业" class="headerlink" title="🚩第5次作业"></a>🚩第5次作业</h2><hr>
<p><strong>3.3.2020作业：第1章 28，29，30，31，32，33，34；第2章 1，2</strong></p>
<ul>
<li><p>第一章</p>
<blockquote>
<p><em>贝叶斯定理（Bayes’ theorem）</em>：</p>
<p>贝叶斯定理是关于<a href="https://baike.baidu.com/item/随机" target="_blank" rel="noopener">随机</a>事件A和B的<a href="https://baike.baidu.com/item/条件概率/4475278" target="_blank" rel="noopener">条件概率</a>（或<a href="https://baike.baidu.com/item/边缘概率/2572198" target="_blank" rel="noopener">边缘概率</a>）的一则定理。其中<code>P(A|B)</code>是在B发生的情况下A发生的可能性。</p>
<p>贝叶斯定理也称贝叶斯推理，早在18世纪，英国学者<a href="https://baike.baidu.com/item/贝叶斯/1405899" target="_blank" rel="noopener">贝叶斯</a>(1702～1763)曾提出计算条件概率的公式用来解决如下一类问题：假设H[1],H[2]…,H[n]互斥且构成一个完全事件，已知它们的概率P(H[i]),i=1,2,…,n,现观察到某事件A与H[1],H[2]…,H[n]相伴随机出现，且已知条件概率P(A|H[i])，求P(H[i]|A)。</p>
<p><em>贝叶斯公式</em>：P(A|B)=P(A)P(B|A)/P(B)</p>
</blockquote>
</li>
</ul>
<hr>
<p>  T_28-34:<img src="https://i.loli.net/2020/03/06/lotecIq9av3TAMp.png" alt=""></p>
<p>  <img src="https://i.loli.net/2020/03/06/BGdMPXA9uhi7wS4.png" alt=""></p>
<hr>
<p>$$<br>  \begin{align}<br>  28.       \<br>  &amp;解:设A = {产品确为合格品}；B = {产品被认为是合格品}\<br>  &amp;由贝叶斯公式得，\&amp;P(A|B)\&amp;= {P(AB)\over P(B)}\&amp;={P(A)P(B|A)\over P(A)P(B|A)+P(\overline A)P(B|\overline A)}\<br>  &amp;= {0.96\times 0.98\over 0.96\times 0.98 + 0.04\times 0.05}\<br>  &amp;= 0.998<br>  \end{align}<br>$$</p>
<hr>
<p>$$<br>\begin{align}<br>29.\<br>解：设&amp;A={客户是谨慎的};B={客户是一般的};\<br>&amp;C={客户是冒失的};D={此客户在一年内出了事故}\<br>&amp;由贝叶斯公式得\<br>P(A|D)&amp;={P(AD)\over P(D)}\<br>&amp;={P(A)P(D|A)\over P(A)P(D|A)+P(B)P(D|B)+P(C)P(D|C)}\<br>&amp;={0.2\times 0.05\over 0.2\times 0.05+0.5\times0.15+0.3\times 0.3}\<br>&amp;=0.057<br>\end{align}<br>$$</p>
<hr>
<p>$$<br>\begin{align}<br>30.\<br>解：设A_i&amp;={第i道工序出次品},(i=1,2,3,4);\<br>P&amp;=1-P(\overline A_1\overline A_2\overline A_3\overline A_4)\<br>&amp;=1-P(\overline A_1)P(\overline A_2)P(\overline A_3)P(\overline A_4)\<br>&amp;=1-0.98\times 0.97\times 0.95\times 0.97\<br>&amp;=0.124<br>\end{align}<br>$$</p>
<hr>
<p>$$<br>\begin{align}<br>31.\<br>解:&amp;设必须经历n次射击，1-(0.8)^n\ge 0.9\<br>&amp;即(0.8)^n\le 0.1;n\ge {1\over lg8}=11.07;至少进行11次射击。<br>\end{align}<br>$$</p>
<hr>
<p>$$<br>\begin{align}<br>32.\<br>证明：&amp;P(A|B)=P(A|\overline B)\<br>&amp;即{P(AB)\over P(B)}={P(A\overline B)\over P(\overline B)}\<br>&amp;P(AB)P(\overline B)=P(A\overline B)P(B)\<br>&amp;P(AB)[1-P(B)]=[P(A)-P(AB)]P(B)\<br>&amp;P(AB)=P(A)P(B);故成立<br>\end{align}<br>$$</p>
<hr>
<p>$$<br>\begin{align}<br>33.\<br>解：&amp;设A_i={第i个能破译},(i=1,2,3)\<br>&amp;P=1-P(\overline A_1\overline A_2\overline A_3)\<br>&amp;P=1-P(\overline A_1)P(\overline A_2)P(\overline A_3)\<br>&amp;P=1-4/5-2/3-3/4=0.6<br>\end{align}<br>$$</p>
<hr>
<p>$$<br>\begin{align}<br>34.\<br>解:&amp;设A={飞机被击落}；B_i={恰有i人击中飞机},(i=1,2,3)\<br>&amp;P(A)=\sum_{i=0}^3P(A|B_i)P(B_i)\<br>&amp;P(A)=(0.4\times 0.5\times 0.3+0.6\times 0.5\times 0.3+0.6\times 0.5\times 0.7)\times 0.2+\&amp;(0.4\times 0.5\times 0.3+0.4\times 0.5\times 0.7+0.6\times 0.5\times 0.7)\times 0.6+0.4\times 0.5\times 0.7\times 1\<br>&amp;=0.458<br>\end{align}<br>$$</p>
<hr>
<ul>
<li>第二章</li>
</ul>
<p>T_1,2:</p>
<p><img src="https://i.loli.net/2020/03/06/w6nAeO8WJ7lkvbG.png" alt="T_1-2.png"></p>
<hr>
<p>$$<br>\begin{align}<br>1.解：&amp;X的可能取值为3，4，5，不同取值的概率为\<br>&amp;P(X=3)={1\over C_5^3}=0.1\<br>&amp;P(X=4)={3\over C_5^3}=0.3\<br>&amp;P(X=5)={C_4^2\over C_5^3}=0.6\</p>
<p>\end{align}<br>$$</p>
<p>故所求X分布率</p>
<table>
<thead>
<tr>
<th>X</th>
<th align="center">3</th>
<th align="center">4</th>
<th align="center">5</th>
</tr>
</thead>
<tbody><tr>
<td>P</td>
<td align="center">0.1</td>
<td align="center">0.3</td>
<td align="center">0.6</td>
</tr>
</tbody></table>
<hr>
<p>$$<br>\begin{align}<br>2.解：&amp;(1)X=0,1,2;\<br>&amp;P(X=0)={C_{13}^3\over C_{15}^3}={22\over 35}\<br>&amp;P(X=1)={C_2^11C_{13}^2\over C_{15}^3}={12\over 35}\<br>&amp;P(X=2)={C_{13}^1\over C_{15}^3}={1\over 35}\<br>\end{align}<br>$$</p>
<p>故X的分布率</p>
<table>
<thead>
<tr>
<th align="center">X</th>
<th align="center">0</th>
<th align="center">1</th>
<th align="center">2</th>
</tr>
</thead>
<tbody><tr>
<td align="center">P</td>
<td align="center">22/35</td>
<td align="center">12/35</td>
<td align="center">1/35</td>
</tr>
</tbody></table>
<p>$$<br>\begin{align}<br>(2)<br>&amp;X的分布函数\<br>&amp;f(n) =<br>\begin{cases}<br>0,  &amp; \text{x&lt;0} \<br>{22\over 35}, &amp; \text{0$\le$x&lt;1}\<br>{34\over 35}, &amp; \text{1$\le$x&lt;2}\<br>1, &amp; \text{x$\ge$2}<br>\end{cases}\<br>(3)&amp;\<br>&amp;P(X\le{1 \over2})=f({1\over 2})={22\over 35}\<br>&amp;P(1&lt;X\le{3 \over2})=f({3\over 2})-f({1\over 2})=0\<br>&amp;P(1\le X\le{3 \over2})=P(X=1)+P(1&lt;X\le{3 \over2})={12\over 35}\<br>&amp;P(1&lt;X&lt;2)=f(2)-f(1)-P(X=2)=0\<br>\end{align}<br>$$</p>
<hr>
<h2 id="🚩第6次作业"><a href="#🚩第6次作业" class="headerlink" title="🚩第6次作业"></a>🚩第6次作业</h2><hr>
<p>习题2：4,5,6,8, 12,14，16，19，24，25</p>
<p><img src="https://i.loli.net/2020/03/15/xPQcBCmFXqudKes.png" alt="4.png"><br>$$<br>\begin{align}<br>解：(1)&amp;由分布律的性质\<br>&amp;1=\sum_{k=0}^∞P(X=k)=a\sum_{k=0}^∞\frac{\lambda^k}{k!}\<br>&amp;a=e^{-\lambda}\<br>(2)&amp;1=\sum_{k=1}^NP(X=k)=a\sum_{k=1}^∞\frac{a}{N}=a\<br>&amp;a=1\<br>\end{align}<br>$$</p>
<p><img src="https://i.loli.net/2020/03/15/BlVCD4KTF8rf6IY.png" alt="5.png"></p>
<p><img src="https://i.loli.net/2020/03/15/LEvrODIxVJnmdjK.jpg" alt="A5.jpg"></p>
<p><img src="https://i.loli.net/2020/03/15/rvA7pLoSWEiOMaZ.png" alt="6.png"></p>
<p><img src="https://i.loli.net/2020/03/15/zEcLZNTIS3q6UMn.jpg" alt="A6.jpg"></p>
<p><img src="https://i.loli.net/2020/03/15/VerH6j9PbkZAIfq.png" alt="8.png"></p>
<p><img src="https://i.loli.net/2020/03/15/QFbRYrn8v1Aai5c.jpg" alt="A8.jpg"></p>
<p><img src="https://i.loli.net/2020/03/15/MdBKufHN7hAcU61.png" alt="12.png"></p>
<p><img src="https://i.loli.net/2020/03/15/oDX13UI8cPvJed2.jpg" alt="A12.jpg"></p>
<p><img src="https://i.loli.net/2020/03/15/SZbRDvnjQUkcY6I.png" alt="14.png"></p>
<p><img src="https://i.loli.net/2020/03/15/EYbVZ2rWy7pOvaJ.jpg" alt="A14.jpg"></p>
<p><img src="https://i.loli.net/2020/03/15/x2cOG4BMISDsE1y.png" alt="16.png"></p>
<p><img src="https://i.loli.net/2020/03/15/BDTviG8elR9fh7F.jpg" alt="A16.jpg"></p>
<p><img src="https://i.loli.net/2020/03/15/V7CimSlxGQ3X1sq.png" alt="19.png"></p>
<p><img src="https://i.loli.net/2020/03/15/ylCmxI9bBrGtshE.jpg" alt="A19.jpg"></p>
<p><img src="https://i.loli.net/2020/03/15/3gxrFWBSRCEjqmL.png" alt="24.png"></p>
<p><img src="https://i.loli.net/2020/03/15/3SnmRzGvbst65ux.jpg" alt="A24.jpg"></p>
<p><img src="https://i.loli.net/2020/03/15/NQZBA41JiMKgLrT.png" alt="25.png"></p>
<p><img src="https://i.loli.net/2020/03/15/uboSB7U3fXZsODJ.jpg" alt="A25.jpg"></p>
<hr>
<h2 id="🚩第7次作业"><a href="#🚩第7次作业" class="headerlink" title="🚩第7次作业"></a>🚩第7次作业</h2><hr>
<p>习题二：20，21，28，29，32</p>
<p><img src="https://i.loli.net/2020/03/21/TkLAhjRgUZW21sY.png" alt="20.png"></p>
<p><img src="https://i.loli.net/2020/03/21/LmJArIdRGH1KceX.jpg" alt="20.jpg"></p>
<p><img src="https://i.loli.net/2020/03/21/8LTyCPB2H9hQ3Al.png" alt="21.png"></p>
<p><img src="https://i.loli.net/2020/03/21/rp9FJwEVMubotkI.jpg" alt="21-1.jpg"></p>
<p><img src="https://i.loli.net/2020/03/21/mbAsfx1cSIUDtC5.jpg" alt="21-2.jpg"></p>
<p><img src="https://i.loli.net/2020/03/21/GIymTv9sU1PLJO2.png" alt="28-29.png"></p>
<p><img src="https://i.loli.net/2020/03/21/Jz6oPda4USZHAnR.jpg" alt="28.jpg"></p>
<p><img src="https://i.loli.net/2020/03/21/K9hSE13lTdiGAO6.jpg" alt="29.jpg"></p>
<p><img src="https://i.loli.net/2020/03/21/MBrNpyUoLvg1Gl5.png" alt="32.png"></p>
<p><img src="https://i.loli.net/2020/03/21/kaVJgBQlLfeSGKZ.jpg" alt="32.jpg"></p>

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